Question

$ConsiderthesetS=\{1,\omega ,{\omega }^2\}$,
where $\omega$ and ${\omega }^2$ are cube roots of unity. If $\ast$ denotes the multiplication operation, the
$structure\{S,\ast \}forms$

• A. A field
• B. A group
• C. A ring
• D. An integral domain

Answer 1

The correct option is B A group

$Thestructure\left(\right\{n,n^{th}\text{roots of unity}\},\times )$ always forms a group. When n=3
$wegettheset\left(\right\{1,w,w^2\},\times )$
which must also form a group.
or
$Thefactthat\left(\right\{1,w,w^2\},\times )formsa$
group can also be seen by the fact that is satisfies the four group properties.

1. Closure:
$\begin{array}{cccc}\ast & 1& w& w^2\\ 1& 1& w& w^2\\ w& w& w^2& 1\\ w^2& w^2& 1& w\end{array}$
From above operation table, we can see that the given operation is closed, on the given set.

2. Associative: "$\ast$" is an associative operation.

3. Identity: From operation table, we can see that the identity element, is "1".

4. Inverse: From operation table we can see that inverse of 1 is 1, inverse of $w$ is $w^2$ and inverse of $w^2$ is $w$.
$So,\left(\right\{1,w,w^2\},\ast )isagroup.$
To be ring, integral domain or field, we need two binary operations to be specified, whereas here we have only one operation given. So, choice (a) is correct.